Folia Mathematica
Vol. 17, No. 1, pp. 3–10 2010 for University of Łódź Pressc
SOME NON-MEASURABLE SETS
ALICJA KIERUS
Abstract. This paper contains constructions of some non-measurable sets, based on classical Vitali’s and Bernstein’s constructions (see for example [6]). This constructions probably belong to mathematical folklore, but as far as we know they are rather hard to be found in literature. It seems that the constructed sets can be used as examples in some interesting situations. 2000 Mathematics Subject Classification: 28A05.
Key words and phrases: Bernstein set, Vitali set, inner Lebesgue measure, Steinhaus property, Hashimoto topology, Density topology.
1. Basic notations and facts
We use standard set theoretic notation. By Q, R we denote, as usuall, the sets of rationals and reals, respectively. We say that C is a Cantor set if it is homeomorphic with the Cantor cube {0, 1}ω. Several times we will use the well-known fact that any interval in R contains a Cantor set of positive Lebesgue measure. By λ we denote Lebesgue measure on R. By λ∗ and λ∗
we denote outer and inner Lebesgue measures, respectively.
Hashimoto topology is a topology on R where any open set U is of the form U = Q\N , where Q is open in natural topology and N is a nullset.
Let A ⊆ R be a measurable set. Put Φ (A) = x ∈ R : lim h→∞ λ (E ∩ [x − h, x + h]) 2h = 1 . We say that a set A ⊆ R is open in density topology if A ⊆ Φ (A).
We say that an operator Ψ : L → P (R), where L is a sigma algebra of all Lebesgue measurable sets, is a density operator if:
i) Ψ (∅) = ∅, Ψ (R) = R,
ii) ∀A,B∈L Ψ (A ∩ B) = Ψ (A) ∩ Ψ (B),
iii) ∀A,B∈L λ (A 4 B) = 0 =⇒ Ψ (A) = Ψ (B),
iv) ∀A∈L λ (A 4 Ψ (A)) = 0.
2. Some modification of the Bernstein set
The Bernstein set is non-measurable in every interval. It has also inner measure zero and full outer measure in every subset. It is such called satu-rated set. We will construct a set that is non-measurable in every interval,
but with positive inner measure in every interval. Also its complement will have the same property.
Lemma 1. There exist Borel pairwise disjoint sets A, B, C ⊆ R such that: ∀(a,b)⊆R (λ ((a, b) ∩ A) > 0 ∧ λ ((a, b) ∩ B) > 0) ∧ λ ((a, b) ∩ C) > 0. Proof. Let {In: n ∈ ω} be an enumeration of open intervals with rational endpoints. We will define inductively families {Cn: n ∈ ω}, {Cn0 : n ∈ ω}
and {Cn00: n ∈ ω} such that:
(1) Cn, Cn0 and Cn00 are pairwise disjoint for any n ∈ ω.
(2) (∀n∈ω) λ In∩S s≤n Cs > 0∧λ In∩S s≤n Cs0 > 0∧λ In∩S s≤n Cs00 > 0 , (3) ∀n∈ω Cn, Cn0 and Cn00 are Borel, nowhere dense sets.
Then A = S n∈ω Cn, B = S n∈ω Cn0 and C = S n∈ω
Cn00 will satisfy the thesis. Indeed, let (a, b) ⊆ R. Then there exist k ∈ ω such that Ik ⊆ (a, b). We
have by (2) λ ((a, b) ∩ A) ≥ λ (Ik∩ A) ≥ λ Ik∩ [ n≤k Cn ! > 0. Analogically λ ((a, b) ∩ B) > 0 and λ ((a, b) ∩ C) > 0.
Let C0 be a Cantor set such that C0 ⊆ I0 and λ (C0) > 0. Since Cantor
set is nowhere dense, there exists an interval (a0, b0) ⊆ I0\C0. Let C00 be
a Cantor set of positive Lebesgue measure such that C00 ⊆ (a0, b0). There
exists an interval (a00, b00) ⊆ (a0, b0) \C00. Let C000 be a Cantor set such that
C000 ⊆ (a0
0, b00), λ (C000) > 0.
Let n ∈ ω. Assume that we have defined Ck, Ck0, Ck00 for all k ≤ n. Since
S
k≤nCk∪Sk≤nCk0 ∪
S
k≤nCk00 is a nowhere dense set, as a finite sum of
nowhere dense sets, there exists an interval (an+1, bn+1) ⊆ In+1\ [ k≤n Ck∪ [ k≤n Ck0 ∪ [ k≤n Ck00 ! . If λIn+1∩Sk≤nCk
> 0, then we put Cn+1 = ∅. Otherwise, let Cn+1 be
a Cantor set such that Cn+1 ⊆ (an+1, bn+1) and λ (Cn+1) > 0. If λ
In+1∩ S k≤nCk0 > 0, then we put Cn+10 = ∅. If λIn+1∩Sk≤nCk0 = 0, then there exists an interval
a0n+1, b0n+1 ⊆ In+1\ [ k≤n+1 Ck∪ [ k≤n Ck0 ∪ [ k≤n Ck00 ! .
In this case we put Cn+10 as a Cantor set such that Cn+10 ⊆ a0 n+1, b0n+1 and λ Cn+10 > 0. If λIn+1∩Sk≤nCk00 > 0, then we put Cn+100 = ∅. If λIn+1∩Sk≤nCk00
= 0, then there exists an interval
a00n+1, b00n+1 ⊆ In+1\ [ k≤n+1 Ck∪ [ k≤n+1 Ck0 ∪ [ k≤n Ck00 ! .
Then we define Cn+100 as a Cantor set such that Cn+100 ⊆ a00
n+1, b00n+1 and
λ Cn+100 > 0. Then sets C0,C1,. . .,Cn+1, C00,C10,. . .,Cn+10 and C000, C100,. . .,Cn+100
obviously satisfy conditions (1) − (3).
Proposition 1. There exists a non-measurable set E ⊆ R such that λ∗((a, b) ∩ E) > 0, λ∗((a, b) ∩ R\E) > 0
and (a, b) ∩ E is non-measurable for all (a, b) ⊆ R.
Proof. Let A, B, C be like in lemma 1. Denote c = |R|. Let F = {Fξ : ξ < c}
be an enumeration of closed subsets of C in subspace topology such that λC(F ) > 0 where λC(F ) = λ (F ∩ C). Let F ∈ F , then there exists
a closed set F0 ⊆ R with F = F0 ∩ C. Since C is a Borel set F is Borel. Moreover, |F | = c since F is a Borel set of positive measure. We will define by induction two disjoint sets {aξ: ξ < c} and {bξ: ξ < c} such that
F ∩ {aξ: ξ < c} 6= ∅ 6= F ∩ {bξ : ξ < c}
for all F ∈ F .
Let a0, b0 be two different elements of F0, and let a1, b1 be two different elements of F1\ {a0, b0}. Assume that aα, bα are defined for all α < β < c.
Then Fβ\ S α<β{aα, bα} = c because S α<β{aα, bα} < c. Hence we can choose two different elements aβ, bβ ∈ Fβ\Sα<β{aα, bα}.
Put E0 = {aξ: ξ < c}. Let F be a closed set, such that F ⊆ E0 ⊆ C.
Therefore F = F ∩C is closed in C. Then λC(F ) = 0. Indeed, if λC(F ) > 0,
then by the construction there exists ξ < c such that bξ ∈ R\E0 ∩ F . It
follows, that F ∩ (R\E0) 6= ∅, so F * E0. We have λC(F ) = λ (F ∩ C) = λ (F ) = 0. Hence
λ∗ E0 = sup {λ (F ) : F ⊂ E and F is closed} = 0.
It follows, that λ∗(E0∩ (a, b)) = 0 for all (a, b) ⊆ R. We will show that
λ∗(E0∩ (a, b)) > 0 for all (a, b) ⊆ R. It will follow that E0is non-measurable
and that for all (a, b) ⊆ R set (a, b) ∩ E0 is non-measurable. Let (a, b) be an open interval and G be an open set such that E0 ∩ (a, b) ⊆ G. Then G0 = G ∩ (a, b) ∩ C is an open set in C. Then C\G0 is a closed in C
subset of E0. It follows that λC(C\G0) = 0. Indeed, if λC(C\G0) > 0 then R\G0∩ E 6= ∅. We have λ (C) = λC(C) = λC C\G0 + λC G0 = 0 + λC G0 , so λC(G0) = λ (C). Then λ (G) ≥ λ (G ∩ (a, b) ∩ C) ≥ λ G0 = λ G0∩ C = λ C G0 = λ (C) . It follows that
λ∗ E0 = inf λ (G) : E0 ⊆ G and G is open ≥ λ (C) .
Therefore E0 is non-measurable in every interval.
Put E = E0∪ A. Then E ∩ (a, b) is non-measurable for any open interval, as a disjoint union of Borel and non-measurable set. Let (a, b) ⊆ R. Then λ∗((a, b) ∩ E) = λ∗ (a, b) ∩ A ∪ E0 ≥ λ∗((a, b) ∩ A) = λ ((a, b) ∩ A) > 0.
It is obvious that B ⊆ R\E. Hence
λ∗((a, b) ∩ (R\E)) ≥ λ∗((a, b) ∩ B) = λ ((a, b) ∩ B) > 0.
The sets constructed in lemma 1 and in proposition 1 are examples of sets which have an empty interior in the Hashimoto ([2]) topology and non-empty interior in every interval (a, b) in density topology ([8]). Indeed, let A, B be sets constructed in lemma 1, and let (a, b) be any interval. Then λ ((a, b) ∩ B) > 0 and (a, b) ∩ A ∩ B = ∅. Therefore A has empty interior in the Hashimoto topology. A has non-empty interior in every interval (a, b) in density topology, because for every interval holds λ ((a, b) ∩ A) > 0.
Each of the sets constructed in lemma 1 is also an example proving that ψ (A) = {x ∈ R : ∀h>0 λ ((x − h, x + h) ∩ A) > 0}
is not a density operator. Indeed, let A, B be sets constructed in lemma 1. Then ψ (A) = R, because for every x ∈ R and every h > 0 we have λ ((x − h, x + h) ∩ A) > 0. But λ (ψ (A) 4 A) = λ (ψ (A) \A) ≥ λ (B\A) = λ (B) > 0.
3. Some modification of the Vitali set
It is known, that for every set, which has a positive measure or is of second category with a Baire property, we have:
This is the Steinhaus or Picard theorem respectively ([7], [5]). Many gener-alizations of these theorems are known (see f.e. [1], [3], [4]). Often instead of (∗) authors write:
(**) int (A − A) 6= ∅.
Let hA, Ii be an algebra A with an ideal I ⊂ A, (X, +, τ ) a topological group. We say that hA, I, τ i has the classical Steinhaus property if
∀A∈A\I 0 ∈ int (A − A) .
We say that hA, I, τ i has the week classical Steinhaus property if ∀A∈A\I int (A − A) 6= ∅.
The set constructed below shows that without any additional assump-tions, the properties (∗) and (∗∗) are not equivalent. Indeed, let A be the set constructed below and A = {R, ∅, A, R\A}, I = {∅}, τ = natural topol-ogy. Then hA, I, τ i has the weak classical Steinhaus property and A does not have the classical Steinhaus property.
Theorem 1. There exists set A ⊆ R such that Int (A − A) 6= ∅ and 0 /∈ Int (A − A).
Proof. Let V be a Vitali non-measurable set constructed in [0, 1] such that 0 ∈ V . Let {an}∞n=1 be an enumeration of all rational numbers in [−1, 1].
We will construct a sequence {Vn}∞n=0 of translated Vitali sets such that
A =S∞
n=0Vn will satisfy the thesis.
Let V0 = V and Vn= ( V + 2n when n is even V + 2n + an+1 2 when n is odd for n ≥ 1.
We will show that (A − A) ∩ (−1, 1) ⊆ (R\Q) ∪ {0}. It will follow that 0 /∈ Int (A − A). Let x ∈ (A − A) ∩ (−1, 1). Then x = v0
1 − v20 for some
v10, v02 ∈ A. It is easy to verify that v10 ∈ Vn, and either v02 ∈ Vn+1, or
v20 ∈ Vn−1, for some n ∈ ω. In any other case, |v10 − v0
2| > 1 which contradicts
the choice of x.
Assume v10 ∈ Vn, v02 ∈ Vn+1 and n is an even number. Let v01 = v1+ 2n
and v20 = v2+ 2(n + 1) + an+2
2 for some v1, v2 ∈ V . Then
v10 − v20 = v1+ 2n − (v2+ 2n + 2 + an+2 2 ) = v1 − v2− 2 − an+2 2 ≤ ≤ v1− v2− 2 + 1 = v1− v2− 1.
Since x ∈ (−1, −1), so v1− v2 > 0. It implies that v1− v2 ∈ R\Q. Hence
Now we will show that [2, 3] ⊆ A − A. Let v0 ∈ Vn where n is odd. Then v0 = v + 2n + an+1
2
for some v ∈ V . Since 0 ∈ V so 2 (n − 1) ∈ Vn−1. Then
v0− 2 (n − 1) ∈ A − A, and
v0− 2 (n − 1) = v + 2n + an+1 2
− 2n + 2 = v + an+1 2 + 2.
It follows that v+an+1
2 +2 ∈ A−A for all v ∈ V . Hence V +a n+1
2 +2 ⊆ A−A.
Similarly V + an+ 2 ⊆ A − A for all n ∈ ω. It implies that ∞ [ n=1 (V + an+ 2) ⊆ A − A and ∞ [ n=1 (V + an+ 2) = 2 + ∞ [ n=1 (V + an) = 2 + [ q∈Q∩[−1,1] (V + q) ⊇ 2 + [0, 1] .
Therefore [2, 3] ⊆ A − A, and (2, 3) = Int ([2, 3]) ⊆ Int (A − A).
Problem 1. We do not know whether there exists a Borel set with this prop-erty.
Acknowledgements The author is gratefully indebted to Professors Marek Balcerzak, Artur Bartoszewicz and Szymon Głąb from Technical University of Lodz for their helpful comments and suggestions.
References
[1] M. Balcerzak, E. Kotlicka, Steinhaus property for products of ideals, Publ. Math. Debrecen 63, 1-2 (2003), 235-248.
[2] H. Hashimoto, On the ∗topology and its application, Fund. Math. 91 (1976), pp. 5-10. [3] Z. Kominek, On an equivalent form of a Steinhaus’s theorem, Mathematica (Cluj) 30
(53), 1 (1988), pp. 25-27.
[4] M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities, PWN, Warszawa-Katowice, 1985.
[5] S. Piccard, Sur les ensembles de distances de ensambles de points d’u espace Euclidean, Mem. Univ. Neuchatel 13 (1939).
[6] J. C. Oxtoby, Measure and Category, Springer-Verlag, Berlin, 1987.
[7] H. Steinhaus, Sur les distances des points des ensambles de measure positive, Fund. Math. 1 (1920), 93-104.
[8] W. Wilczyński, Density topologies, Handbook of Measure Theory, Ed. E. Pap. Elsevier, chapter 15 (2002), 675-702.
Alicja Kierus
Institute of Mathematics, Technical University of Łódź Wólczańska 215, 93-005 Łódź, Poland
